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Capacitance

166007 Calculate charge on capacitor in steady state.

1

50 μ C

2

30 μ C

3

45 μ C

4

60 μ C

Explanation:

: In steady state the capacitor becomes open, so current does not flow through the capacitor.
Then current flow through the resistance in the given circuit.

Current, $i = \frac{V}{R_{eq}} = \frac{9}{27 \times 10^{3}}$
Voltage across the capacitor $(V_{c}) = i \times R_{1}$
$= \frac{9}{27 \times 10^{3}} \times 15 \times 10^{3} = 5 V$
Charge through the capacitor $q = {CV}_{e}$
$= 9 \times 10^{- 6} \times 5$
$= 45 \times 10^{- 6} C = 45 μ C$

[AIIMS-25.05.2019(E)]**# Shift-2]

Capacitance

166008 What is the energy stored in a $50 mH$ inductor carrying a current of $4 A$ ?

1

0.4 J

2

0.2 J

3

0.05 J

4

0.1 J

Explanation:

: Given,
Inductance, $L = 50 mH$
Current, $I = 4 A$
Energy stored in an inductor $= \frac{1}{2} {LI}^{2}$
$= \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$= 50 \times 10^{- 3} \times 8$
$= 0.4 J$

[JCECE-2004]

Capacitance

166009 In the adjoining figure, $E = 5 V, r = 1 Ω, R_{2} =$ $4 Ω, R_{1} = R_{3} = 1 Ω$ and $C = 3 μ F$ . The numerical value of the charge on each plate of the capacitor is

1

3 μ C

2

6 μ C

3

12 μ C

4

24 μ C

Explanation:

: At steady state condition there is no current flow through capacitor.
So, $I = \frac{E}{R_{2} + r} = \frac{5}{4 + 1} = 1 A$
Potential difference across capacitor
$= I \times R_{2}$
$= 1 \times 4 = 4 V$
Charge on each capacitor $= (\frac{C}{2}) \times 4$
$= \frac{3}{2} \times 4 = 6 μ C$

[JCECE-2013]

Capacitance

166010 A voltage $V_{P Q} = V_{0} \cos ω t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c = \frac{1}{ω R \sqrt{3}} and L = \frac{R \sqrt{3}}{ω}$
Then, the total impedance between $P$ and $Q$ is

1

1.5 R

2

2 R

3

3 R

4

4 R

5

2.5 R

Explanation:

: Given that,
Capacitance, $C = \frac{1}{ω R \sqrt{3}}$ ,
Inductance, $L = \frac{R \sqrt{3}}{ω}$

$Z_{1} = R + j ω L = R + j ω (\frac{R \sqrt{3}}{ω}) = R + j R \sqrt{3}$
$Z_{2} = R - j \frac{1}{ω C} = R - j \frac{1}{ω (\frac{1}{ω R \sqrt{3}})}$
$= R - j R \sqrt{3}$
$∵$ Impedance $Z_{1}$ and $Z_{2}$ are in parallel so,
$Z_{eq} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}}$
$= \frac{(R + j R \sqrt{3}) \cdot (R - j R \sqrt{3})}{(R + j R \sqrt{3}) + (R - j R \sqrt{3})}$
$= \frac{R^{2} + R^{2} (\sqrt{3})^{2}}{2 R}$
$Z_{eq} = \frac{4 R^{2}}{2 R} = 2 R$
So, total impedance between $P$ and $Q$ is
$Z_{PQ} = R + Z_{eq} = R + 2 R = 3 R$

[Kerala CEE -2018]

Capacitance

166007 Calculate charge on capacitor in steady state.

1

50 μ C

2

30 μ C

3

45 μ C

4

60 μ C

Explanation:

: In steady state the capacitor becomes open, so current does not flow through the capacitor.
Then current flow through the resistance in the given circuit.

Current, $i = \frac{V}{R_{eq}} = \frac{9}{27 \times 10^{3}}$
Voltage across the capacitor $(V_{c}) = i \times R_{1}$
$= \frac{9}{27 \times 10^{3}} \times 15 \times 10^{3} = 5 V$
Charge through the capacitor $q = {CV}_{e}$
$= 9 \times 10^{- 6} \times 5$
$= 45 \times 10^{- 6} C = 45 μ C$

[AIIMS-25.05.2019(E)]**# Shift-2]

Capacitance

166008 What is the energy stored in a $50 mH$ inductor carrying a current of $4 A$ ?

1

0.4 J

2

0.2 J

3

0.05 J

4

0.1 J

Explanation:

: Given,
Inductance, $L = 50 mH$
Current, $I = 4 A$
Energy stored in an inductor $= \frac{1}{2} {LI}^{2}$
$= \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$= 50 \times 10^{- 3} \times 8$
$= 0.4 J$

[JCECE-2004]

Capacitance

166009 In the adjoining figure, $E = 5 V, r = 1 Ω, R_{2} =$ $4 Ω, R_{1} = R_{3} = 1 Ω$ and $C = 3 μ F$ . The numerical value of the charge on each plate of the capacitor is

1

3 μ C

2

6 μ C

3

12 μ C

4

24 μ C

Explanation:

: At steady state condition there is no current flow through capacitor.
So, $I = \frac{E}{R_{2} + r} = \frac{5}{4 + 1} = 1 A$
Potential difference across capacitor
$= I \times R_{2}$
$= 1 \times 4 = 4 V$
Charge on each capacitor $= (\frac{C}{2}) \times 4$
$= \frac{3}{2} \times 4 = 6 μ C$

[JCECE-2013]

Capacitance

166010 A voltage $V_{P Q} = V_{0} \cos ω t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c = \frac{1}{ω R \sqrt{3}} and L = \frac{R \sqrt{3}}{ω}$
Then, the total impedance between $P$ and $Q$ is

1

1.5 R

2

2 R

3

3 R

4

4 R

5

2.5 R

Explanation:

: Given that,
Capacitance, $C = \frac{1}{ω R \sqrt{3}}$ ,
Inductance, $L = \frac{R \sqrt{3}}{ω}$

$Z_{1} = R + j ω L = R + j ω (\frac{R \sqrt{3}}{ω}) = R + j R \sqrt{3}$
$Z_{2} = R - j \frac{1}{ω C} = R - j \frac{1}{ω (\frac{1}{ω R \sqrt{3}})}$
$= R - j R \sqrt{3}$
$∵$ Impedance $Z_{1}$ and $Z_{2}$ are in parallel so,
$Z_{eq} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}}$
$= \frac{(R + j R \sqrt{3}) \cdot (R - j R \sqrt{3})}{(R + j R \sqrt{3}) + (R - j R \sqrt{3})}$
$= \frac{R^{2} + R^{2} (\sqrt{3})^{2}}{2 R}$
$Z_{eq} = \frac{4 R^{2}}{2 R} = 2 R$
So, total impedance between $P$ and $Q$ is
$Z_{PQ} = R + Z_{eq} = R + 2 R = 3 R$

[Kerala CEE -2018]

Capacitance

166007 Calculate charge on capacitor in steady state.

1

50 μ C

2

30 μ C

3

45 μ C

4

60 μ C

Explanation:

: In steady state the capacitor becomes open, so current does not flow through the capacitor.
Then current flow through the resistance in the given circuit.

Current, $i = \frac{V}{R_{eq}} = \frac{9}{27 \times 10^{3}}$
Voltage across the capacitor $(V_{c}) = i \times R_{1}$
$= \frac{9}{27 \times 10^{3}} \times 15 \times 10^{3} = 5 V$
Charge through the capacitor $q = {CV}_{e}$
$= 9 \times 10^{- 6} \times 5$
$= 45 \times 10^{- 6} C = 45 μ C$

[AIIMS-25.05.2019(E)]**# Shift-2]

Capacitance

166008 What is the energy stored in a $50 mH$ inductor carrying a current of $4 A$ ?

1

0.4 J

2

0.2 J

3

0.05 J

4

0.1 J

Explanation:

: Given,
Inductance, $L = 50 mH$
Current, $I = 4 A$
Energy stored in an inductor $= \frac{1}{2} {LI}^{2}$
$= \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$= 50 \times 10^{- 3} \times 8$
$= 0.4 J$

[JCECE-2004]

Capacitance

166009 In the adjoining figure, $E = 5 V, r = 1 Ω, R_{2} =$ $4 Ω, R_{1} = R_{3} = 1 Ω$ and $C = 3 μ F$ . The numerical value of the charge on each plate of the capacitor is

1

3 μ C

2

6 μ C

3

12 μ C

4

24 μ C

Explanation:

: At steady state condition there is no current flow through capacitor.
So, $I = \frac{E}{R_{2} + r} = \frac{5}{4 + 1} = 1 A$
Potential difference across capacitor
$= I \times R_{2}$
$= 1 \times 4 = 4 V$
Charge on each capacitor $= (\frac{C}{2}) \times 4$
$= \frac{3}{2} \times 4 = 6 μ C$

[JCECE-2013]

Capacitance

166010 A voltage $V_{P Q} = V_{0} \cos ω t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c = \frac{1}{ω R \sqrt{3}} and L = \frac{R \sqrt{3}}{ω}$
Then, the total impedance between $P$ and $Q$ is

1

1.5 R

2

2 R

3

3 R

4

4 R

5

2.5 R

Explanation:

: Given that,
Capacitance, $C = \frac{1}{ω R \sqrt{3}}$ ,
Inductance, $L = \frac{R \sqrt{3}}{ω}$

$Z_{1} = R + j ω L = R + j ω (\frac{R \sqrt{3}}{ω}) = R + j R \sqrt{3}$
$Z_{2} = R - j \frac{1}{ω C} = R - j \frac{1}{ω (\frac{1}{ω R \sqrt{3}})}$
$= R - j R \sqrt{3}$
$∵$ Impedance $Z_{1}$ and $Z_{2}$ are in parallel so,
$Z_{eq} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}}$
$= \frac{(R + j R \sqrt{3}) \cdot (R - j R \sqrt{3})}{(R + j R \sqrt{3}) + (R - j R \sqrt{3})}$
$= \frac{R^{2} + R^{2} (\sqrt{3})^{2}}{2 R}$
$Z_{eq} = \frac{4 R^{2}}{2 R} = 2 R$
So, total impedance between $P$ and $Q$ is
$Z_{PQ} = R + Z_{eq} = R + 2 R = 3 R$

[Kerala CEE -2018]

Capacitance

166007 Calculate charge on capacitor in steady state.

1

50 μ C

2

30 μ C

3

45 μ C

4

60 μ C

Explanation:

: In steady state the capacitor becomes open, so current does not flow through the capacitor.
Then current flow through the resistance in the given circuit.

Current, $i = \frac{V}{R_{eq}} = \frac{9}{27 \times 10^{3}}$
Voltage across the capacitor $(V_{c}) = i \times R_{1}$
$= \frac{9}{27 \times 10^{3}} \times 15 \times 10^{3} = 5 V$
Charge through the capacitor $q = {CV}_{e}$
$= 9 \times 10^{- 6} \times 5$
$= 45 \times 10^{- 6} C = 45 μ C$

[AIIMS-25.05.2019(E)]**# Shift-2]

Capacitance

166008 What is the energy stored in a $50 mH$ inductor carrying a current of $4 A$ ?

1

0.4 J

2

0.2 J

3

0.05 J

4

0.1 J

Explanation:

: Given,
Inductance, $L = 50 mH$
Current, $I = 4 A$
Energy stored in an inductor $= \frac{1}{2} {LI}^{2}$
$= \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$= 50 \times 10^{- 3} \times 8$
$= 0.4 J$

[JCECE-2004]

Capacitance

166009 In the adjoining figure, $E = 5 V, r = 1 Ω, R_{2} =$ $4 Ω, R_{1} = R_{3} = 1 Ω$ and $C = 3 μ F$ . The numerical value of the charge on each plate of the capacitor is

1

3 μ C

2

6 μ C

3

12 μ C

4

24 μ C

Explanation:

: At steady state condition there is no current flow through capacitor.
So, $I = \frac{E}{R_{2} + r} = \frac{5}{4 + 1} = 1 A$
Potential difference across capacitor
$= I \times R_{2}$
$= 1 \times 4 = 4 V$
Charge on each capacitor $= (\frac{C}{2}) \times 4$
$= \frac{3}{2} \times 4 = 6 μ C$

[JCECE-2013]

Capacitance

166010 A voltage $V_{P Q} = V_{0} \cos ω t$ ( where, $V_{0}$ is a real amplitude) is applied between the points $P$ and $Q$ in the network shown in the figure. The values of capacitance and inductance are
$c = \frac{1}{ω R \sqrt{3}} and L = \frac{R \sqrt{3}}{ω}$
Then, the total impedance between $P$ and $Q$ is

1

1.5 R

2

2 R

3

3 R

4

4 R

5

2.5 R

Explanation:

: Given that,
Capacitance, $C = \frac{1}{ω R \sqrt{3}}$ ,
Inductance, $L = \frac{R \sqrt{3}}{ω}$

$Z_{1} = R + j ω L = R + j ω (\frac{R \sqrt{3}}{ω}) = R + j R \sqrt{3}$
$Z_{2} = R - j \frac{1}{ω C} = R - j \frac{1}{ω (\frac{1}{ω R \sqrt{3}})}$
$= R - j R \sqrt{3}$
$∵$ Impedance $Z_{1}$ and $Z_{2}$ are in parallel so,
$Z_{eq} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}}$
$= \frac{(R + j R \sqrt{3}) \cdot (R - j R \sqrt{3})}{(R + j R \sqrt{3}) + (R - j R \sqrt{3})}$
$= \frac{R^{2} + R^{2} (\sqrt{3})^{2}}{2 R}$
$Z_{eq} = \frac{4 R^{2}}{2 R} = 2 R$
So, total impedance between $P$ and $Q$ is
$Z_{PQ} = R + Z_{eq} = R + 2 R = 3 R$

[Kerala CEE -2018]

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